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Height of football

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data

    A quarterback throws a football with an initial speed of 19.0 m/s at an angle 30.5deg above the horizontal. Unfortunately, the pass is incomplete and the football drops to the ground, 35.0 m away from the quaterback. From what height was the football released? (no diagram provided)

    2. Relevant equations

    Vy= Vknoty - gt from which we get ttot= [tex]\frac{2Vosin\vartheta}{g}[/tex]
    from the equation y= Voyty2-0.5gt0.52 we get hmax= v02sin2g[tex]\vartheta[/tex]/2g
    3. The attempt at a solution
    so to figure out far it goes i used the equation xmax= v0xttot where it wld now equal vknotcos[tex]\vartheta[/tex](2*vknot sin[tex]\vartheta[/tex]/g) + distance it travelled and i wanted to know what i was doing wrong thanx :))
     
  2. jcsd
  3. Jun 6, 2009 #2

    Cyosis

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    Not sure what you're doing here. You set v_y=0 when the ball touches the ground? This is not true for if there had been no ground it would have continued to fall.

    You know the ball traveled a distance of 35 meters in the x-direction. So you can calculate how long it took for the ball to reach that position with [itex]x=v_x t[/itex]. Knowing the flight time you can solve the initial height by using [itex]y=y_0+v_0 t-1/2 gt^2[/itex].
     
  4. Jun 6, 2009 #3
    why wld the velocity be v0 and not Vy??
     
  5. Jun 6, 2009 #4

    Cyosis

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    It's not I just listed the general kinematic expression for a falling object. In your case the vertical starting velocity is v_y.
     
    Last edited: Jun 6, 2009
  6. Jun 6, 2009 #5
    thanx :))
     
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